If you want to write a binary number you need to use the 0b prefix. std::uint64_t a = 0b0000000000000000000000000000000000000000000000001111111100000000; std::bitset<64> b(a); Your example fixed and working live...

"How can I test whether every element in a bitset has same value using gmock and gtest." The AllOf() matcher is meant to combine other matchers as explained in the reference documentation. The single Eq(true) requires to have flags an automatic cast operator for bool (or at least int),...

If you can efficiently query the number of bits set in every range, you can perform binary search on #set_bits(0,i) to find the first index where this value equals k. It will take O(log(n)*f(n)), where f(n) is the complexity of #set_bits(0,i) op....

Assuming you want to start at ASCII code 64, and that 'a' (or 'A') is simply 000001 in that case, then you can simply do c1 = static_cast<char>(std::bitset<6>(arr[0]).to_ulong() + 64); // 'A' in decimal is 65, in binary is 0b01000001. 'a' in decimal is 97, in binary is 0b01100001. In...

If you refer docs of BitSet , there is valueOf() method public static BitSet valueOf(byte[] bytes) Returns a new bit set containing all the bits in the given byte array. BitSet a = BitSet.valueOf(test); ...

That is not an "overhead", that't the marker that lets Java figure out what type it needs to create when deserializing the object from that file. Since ObjectInputStream has no idea what you have serialized into a file, and has no way for you to provide a "hint", ObjectOutputStream must...

I'm one of those who would give you a similar performance answer, but I'll try to give you something a bit more in-depth than "just because". It is something I came across through actual profiling and timing, not merely distrust and paranoia. One of the biggest problems with bitset and...

You can do it by xor-ing the two sets together, and comparing the length of the result to lengths of bit sets: If xor is empty, bit sets are equal. You can bypass this operation by calling equals() Otherwise, the length of xor result will be equal to the position...

BitSet.cardinality() returns the number of bits set to true in the BitSet. I think you are looking for BitSet.size(). But keep in mind it will return the number of bits, not bytes. Assuming after Huffman encoding you have approximately half of the bits set to true, that means your BitSet...

This is undefined behavior. The standard merely states that The class template bitset<N> describes an object that can store a sequence consisting of a fixed number of bits, N. It says nothing about the layout of this class internally. There is no guarantee that sizeof(bitset<8>) is 1. On my implementation,...

This is because the BitSet constructor creates a BitSet "whose initial size is large enough to explicitly represent" bits in the range given by the parameter. So the actual size will be at least the number you give in the parameter, but not necessarily equal to that number. The reason...

Matthew Austern wrote an iterator for bitset here: http://www.drdobbs.com/the-standard-librarian-bitsets-and-bit-v/184401382?pgno=2 It's over 100 lines so I feel just lifting it and putting it in this answer may be a bit out of bounds. But it works marvelously for STL algorithms. Austern answers this exact question: While bitset doesn’t have the STL...

The template argument need to be a full and complete type. And a templated class like std::bitset is not a complete type without its size. So you need to do e.g. std::queue<bitset<8>> buttonQueue; In other words, you need to provide the bitset size as well....

Very few CPUs have builtin instructions for GF(2) division like this, so you'll need to implement it yourself with shifts and xors. Basically, you implement it exactly like you did it on paper -- shift the divisor up until its top bit matches that of dividend, then xor and shift...

As far as I can tell, you basically iterate over each bit. As such, I'd imagine simply shifting and masking off the LSB every time should provide good performance. Something like: uint64_t a = *cursor; for(int bit = 0; a != 0; bit++, a >>= 1) { hist[bit] -= (a...

This answer includes a BitSet subclass that has a toByteArray method. To write, you can get the byte[] array from that method and use ByteBuffer.put(byte[]) (ByteBuffer docs). To read, you can use get() and then loop over the byte[] and rebuild the bitset. (For reference: FileChannel docs) EDIT in answer...

BitSet will be less memory, using only one bit is far more efficient. The method overhead you are looking at is once no matter how many instances of your class you have, so its cost is amortized basically to 0 The advantage of a boolean over an array of booleans...

std::bitset has no range constructor, so you will have to loop, but setting every bit individually with std::bitset::set() is underkill. std::bitset has support for binary operators, so you can at least set 64 bits in bulk: std::bitset<192> bs; for(int i = 2; i >= 0; --i) { bs <<= 64;...

You've created a 9x9 array of BitSet references, but set every element value to the same reference - there's only a single BitSet object involved. This is just a more complex version of this: StringBuilder builder = new StringBuilder(); StringBuilder[] array = { builder, builder }; array[0].append("foo"); System.out.println(array[1]); // Prints...

The code you have is very close. Here are a few thoughts about where it might be off. The bfOffBits value must include the size of the palette. fileHeader.bfOffBits = sizeof(BITMAPFILEHEADER) + (sizeof(BITMAPINFOHEADER)) + 2*sizeof(RGBQUAD); Some software may interpret 0 as white and 1 as black, regardless of what the...

BitSet has a static valueOf(long[]) method which Returns a new bit set containing all the bits in the given long array. So an array with one long will have 64 bits, an array with two longs will have 128 bits, etc. If you only need to get a BitSet from...

The question does allow for alternative solutions. Java's BitSet can work but there are a couple of hidden traps: The Java VM will need some memory. So you may run out of memory. The BitSet is backed by an array. Java arrays use 32bit signed int as indexes, so you...

To answer about mapping 4D to 1D, if you visualize, say, a chess board, you can come up with the formula for 2D to 1D by thinking if every row has width elements, and I first go down row number of rows and then move over to col, then I'm...

The interface doesn't support this directly. You could construct a temporary bitset instead: // assuming op1 is larger op1 |= std::bitset<op1.size()>(op2.to_ullong()) If the larger bitset's value wouldn't fit in a unsigned long long, youl could call to_string instead....